Question

Find the derivative of
$\sf \: {e}^{ log( \sqrt{1 + {tan}^{2} }x ) }$
(a) tan x
(b) sec x
(c) sec^2x
(d) sec(x).tan(x) ​

Answer 1

Answer:

Opt 4) Sec X . Tan X

Step-by-step explanation:

Refer to the attachment

Answer 2

GIVEN :–

$\\{ \bold{\implies y = {e}^{ log \left \{ \sqrt{1 + {tan}^{2} (x)} \: \right\}}}}$

TO FIND :–

$\\{ \bold{\implies \dfrac{dy}{dx} = ? }}$

SOLUTION :–

$\\{ \bold{\implies y = {e}^{ log \left \{ \sqrt{1 + {tan}^{2} (x)} \: \right\}}}}$

• Using log property –

$\\ \dashrightarrow \: \: { \bold{ {e}^{ log(x) = x}}}$

• So that –

$\\{ \bold{\implies y = \sqrt{1 + {tan}^{2} (x)} }}$

• We also know that –

$\\ \: \: \dashrightarrow \: \: { \bold{ 1 + {tan}^{2} (x) = { \sec}^{2}(x) }}$

• So that –

$\\{ \bold{\implies y = \sqrt{{ \sec}^{2}(x)}}}$

$\\{ \bold{\implies y = \sec(x) }}$

• Now difference with respect to 'x' –

$\\{ \bold{\implies \dfrac{dy}{dx} = \dfrac{d \{\sec(x) \}}{dx}}}$

$\\ \left[ { \bold{ \: \: \because \: \: \dfrac{d \{\sec(x) \}}{dx} = \sec(x). \tan(x)}} \right] \:$

$\\{ \bold{\implies \dfrac{dy}{dx} = \sec(x). \tan(x)}}$

$\\ \implies \large { \boxed{ \bold{ \dfrac{dy}{dx} = \sec(x). \tan(x)}}}$

▪︎ Hence, Option (d) is correct.